What are the critical values, if any, of #f(x)= (x^2+6x-7)^2#?

Question

1861 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-19T04:03:47

The critical values for #f# are #-7,-3, "and " 1#

Use the power rule and the chain rule to find:

#f'(x) = 2(x^2+6x-7)(2x+6)#.

Since, #f(x)# and #f'(x)# are polynomials, the critical values for #f# are the zeros of #f'#

Solve #f'(x) = 2(x^2+6x-7)(2x+6)=0# by solving

#x^2+6x-7=0# or #2x+6=0#. So,

#(x+7)(x-1)=0# or #x=-3#

#x=-7, 1, "or" -3#

The critical values for #f# are #-7,-3, "and " 1#